Net change and accumulation from a rate practice problems
Amount now = amount at the start + the integral of the rate of change. 10 problems with worked solutions; in 10 of them every equation is proved by a computer algebra system.
A town's population grows at \( \displaystyle r(t) = - t^{2} + 6 t + 30 \) people per year. The population is 1200 at \( \displaystyle t = 0 \). What is it at \( \displaystyle t = 2 \)?
A particle starts at position 0 m with velocity \( \displaystyle v(t) = 13 \sqrt{t} \) m/s. Where is it at \( \displaystyle t = 4 \)?
A particle starts at position 500 m with velocity \( \displaystyle v(t) = 2 t + 5 \) m/s. Where is it at \( \displaystyle t = 3 \)?
Oil leaks into a containment boom at \( \displaystyle r(t) = 3 t^{2} + 5 \) gallons per hour, starting from 0 gallons. How much has collected at \( \displaystyle t = 6 \) hours?
Water flows into a tank at a rate of \( \displaystyle r(t) = 2 t + 4 \) liters per minute. The tank holds 50 liters at \( \displaystyle t = 0 \). How much water is in the tank at \( \displaystyle t = 3 \)?
A particle starts at position 20 m with velocity \( \displaystyle v(t) = 3 t + 8 \) m/s. Where is it at \( \displaystyle t = 5 \)?
A particle starts at position 50 m with velocity \( \displaystyle v(t) = 2 t + 5 \) m/s. Where is it at \( \displaystyle t = 4 \)?
Oil leaks into a containment boom at \( \displaystyle r(t) = 5 t + 16 \) gallons per hour, starting from 20 gallons. How much has collected at \( \displaystyle t = 3 \) hours?
A town's population grows at \( \displaystyle r(t) = - t^{2} + 6 t + 30 \) people per year. The population is 500 at \( \displaystyle t = 0 \). What is it at \( \displaystyle t = 5 \)?
Oil leaks into a containment boom at \( \displaystyle r(t) = 100 e^{- \frac{t}{2}} \) gallons per hour, starting from 50 gallons. How much has collected at \( \displaystyle t = 2 \) hours?